according to a survey, 67% of households have cable tv, 74% of households have 2 or more tv sets, and 55% of households have cable tv and two or more tv sets. a) what is the probability that the household has cable tv and only one set of tv? b) if the random chosen household has 2 or more tv sets, what is the probability that the household also has cable tv?

Question

ganpat · Answer

probability of cable TV = 0.67
probability of 2 or more TV sets = 0.74
probability of Cable TV and two or more sets = 0.5

ganpat · Answer

a) Probability for that household as cable tv and one set of TV = (1-0.55)
b) Chosen is 2 orr more set, probability of cable TV = (0.74 * 0.67)
I am not very sure.. @ganeshie8  @Hero

mathmate · Answer

Let 
A=event that household has cable
B=event that household has $\ge2$ TV sets
We assume that any household that has cable has at least one TV-set.
Then the following Venn diagram applies
(a)Probability of having cable and NOT two or more TV-sets =>  $P(A\cap ~B)$
(b)Probability of having cable given household has 2+ tv sets.
=> $P(A|B)=P(A\cap B)/P(B)$
Use the following Venn Diagram (of probabilities) to help you.
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mathmate · Answer

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